Projection markovienne de processus stochastiques

tel-00766235, version 1 - 17 Dec 2012
Using the density of Im(λ−A) in Im(λ−A) = C0([0,∞×R d ), we get
∀g ∈ C0([0,∞[×R d ∞
), e −λt ∞
Utg(0,x0)dt = e −λt Ptg(0,x0)dt,
0
(2.20)
so the Laplace transform of t ↦→ Ptg(0,x0) is uniquely determined.
Using (2.18), for any h ∈ D 0 , t ↦→ Pth(0,x0) is right-continuous:
∀h ∈ D 0 , lim
t↓ǫ Pth(0,x0) = Pǫh(0,x0).
Furthermore, the density of D0 in C0([0,∞[×R d ) implies the weakcontinuity
of t → Ptg(0,x0) for any g ∈ C0([0,∞[×R d ). Indeed, let
g ∈ C0([0,∞[×R d ), there exists (hn)n≥0 ∈ D0 such that
Then equation (2.18) yields,
|Ptg(0,x0)−Pǫg(0,x0)|
lim
n→∞ g −hn = 0
= |Pt(g −hn)(0,x0)+(Pt −Pǫ)hn(0,x0)+Pǫ(g −hn)(0,x0)|
≤ |Pt(g −hn)(0,x0)|+|(Pt −Pǫ)hn(0,x0)|+|Pǫ(g −hn)(0,x0)|
≤ 2g −hn+|(Pt −Pǫ)hn(0,x0)|
Using the right-continuity of t ↦→ Pthn(0,x0) for any n ≥ 0, taking t ↓ ǫ
then n → ∞, yields
lim
t↓ǫ Ptg(0,x0) = Pǫg(0,x0).
Thusthetworight-continuousfunctionst ↦→ Ptg(0,x0)andt ↦→ Utg(0,x0)
have the same Laplace transform by (2.20), which implies they are
equal:
∀g ∈ C0([0,∞[×R d
), g(t,y)q0,t(x0,dy) = g(t,y)pt(x0,dy).
(2.21)
By [40, Proposition 4.4, Chapter 3], C0([0,∞[×R d ) is convergence determining,
hence separating, allowing us to conclude that pt(x0,dy) =
q0,t(x0,dy).
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